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2 years ago

A large tuning fork has a lower or higher frequency than a small tuning fork?

Physics

1 answer:

netineya [11]2 years ago

5 0

A large tuning fork makes a lower frequency than a small one

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True or False: Amplitude can increase or decrease wavelength

Marat540 [252]

Answer:

A. False, frequency can increase or decrease wavelength.

For example: a high frequency would mean there are shorter wavelengths that occur in a period. Meanwhile, a low frequency would indicate that the wavelengths are longer and in longer periods.

4 0

2 years ago

An ice boat is coasting along a frozen lake. Friction between the ice and the boat is negligible, and so is air resistance. Noth

zhannawk [14.2K]

Answer:

(A) No

(B) Speed decreases

Explanation:

(A) since there is nothing propelling the boat and the friction between the ice and the boat and also air resistance is negligible the net force of the system in the horizontal direction is zero and hence there is no change in the horizontal momentum of the boat.

(B) Since the person had not velocity in the horizontal direction before landing on the boat but now has one after landing on the boat, the speed of the boat will decrease because the momentum has to be conserved (remember there is no change in it).

6 0

3 years ago

What must be the pressure difference between the two ends of a 2.6 km section of pipe, 36 cm in diameter, if it is to transport

ad-work [718]

Answer: $1.13(10)^{3} Pa$

Explanation:

This problem can be solved by the following equation:

$\Delta P=\frac{8 \eta L Q}{\pi r^{4}}$

Where:

$\Delta P$ is the pressure difference between the two ends of the pipe

$\eta=0.20 Pa.s$ is the viscosity of oil

$L=2.6 km=2600 m$ is the length of the pipe

$Q=900 \frac{cm^{3}}{s} \frac{1 m^{3}}{(100 cm)^{3}}=0.0009 \frac{m^{3}}{s}$ is the Rate of flow of the fluid

$d=36 cm=0.36 m$ is the diameter of the pipe

$r=\frac{d}{2}=0.18 m$ is the radius of the pipe

Soving for $\Delta P$ :

$\Delta P=\frac{8 (0.20 Pa.s)(2600 m)(0.0009 \frac{m^{3}}{s})}{\pi (0.18 m)^{4}}$

Finally:

$\Delta P=1135.26 Pa \approx 1.13(10)^{3} Pa$

7 0

2 years ago

A typical electric refrigerator has a power rating of 500 Watts, which is the rate (J/s) at which electrical energy is supplied

Goshia [24]

Answer:

The rate of heat removed from inside the refrigerator is 300 watts.

Explanation:

By the First Law of Thermodynamics and the definition of a Refrigeration Cycle, we have the following formula to determine the rate of heat removed from inside the refrigerator ( $\dot Q_{L}$ ), in watts: